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## anonymous one year ago I have the answer but i dont know how.. The lower edge of a mural, 12 ft high, is 6ft above an observer’s eye. Under the assumption that the most favorable view is obtained when the angle subtended by the mural at the eye is maximum, at what distance from the wall should the observer stand?

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1. ganeshie8

|dw:1433509816192:dw|

2. anonymous

we have the same drawing sir

3. ganeshie8

From the lower small triangle $\tan(s) = \frac{6}{x}\tag{1}$ From the big triangle $\tan(t+s)=\frac{18}{x}\tag{2}$

4. ganeshie8

Our goal is to eliminate $$s$$ somehow and get an equation with $$t$$ and $$x$$ as variables

5. ganeshie8

In view of that, apply the angle sum formula : $\large \tan(t)=\tan((t+s)-s)=\frac{\tan(t+s)-\tan(s)}{1+\tan(t+s)\tan(s)}$ plugin the values

6. anonymous

ok sir i will solve now

7. anonymous

i got this sir: |dw:1433510419375:dw|

8. ganeshie8

doesn't look correct

9. ganeshie8

\large{\begin{align} \tan(t)=\tan((t+s)-s) &=\frac{\tan(t+s)-\tan(s)}{1+\tan(t+s)\tan(s)}\\~\\ &=\dfrac{\dfrac{18}{x}-\dfrac{6}{x}}{1+\dfrac{18}{x}\dfrac{6}{x}}\\~\\ &=\dfrac{12x}{x^2+108} \end{align}}

10. ganeshie8

so the equation is $\large \tan(t)=\dfrac{12x}{x^2+108}$ and we want to maximize $$t$$

11. anonymous

i got that equation sir

12. ganeshie8

may be isolate $$t$$ and find $$\dfrac{dt}{dx}$$

13. anonymous

is this the bases sir? |dw:1433510949377:dw|

14. ganeshie8

$\large \tan(t)=\dfrac{12x}{x^2+108}$ $\large t = \tan^{-1}\left(\dfrac{12x}{x^2+108}\right)$ $\large \dfrac{dt}{dx} = ?$

15. ganeshie8

remember the derivative of $$\tan^{-1}(x)$$ ?

16. anonymous

|dw:1433511150272:dw|

17. ganeshie8

Yes use that

18. ganeshie8

$\large t = \tan^{-1}\left(\dfrac{12x}{x^2+108}\right)$ \large{\begin{align} \dfrac{dt}{dx} &= \dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\dfrac{d}{dx}\left(\dfrac{12x}{x^2+108}\right)\\~\\~\\ &=? \end{align}}

19. anonymous

ok wait sir i will solve

20. anonymous

i got this sir:

21. anonymous

$\frac{ 1 }{ 1+\frac{ 12X ^{2} }{ x ^{4}+216x ^{2}+11664 } }\times \frac{ -12x ^{2}+1296 }{ x ^{4}+216x ^{2}+11664 }$

22. anonymous

ohh there an error im sorry.. it should be 144x^2 instead of having 12x^2

23. anonymous

and i was stuck into this.. i dont know how to continue...

24. anonymous

i already have an headache..

25. ganeshie8

do not expand

26. ganeshie8

because we will be setting all that equal to 0 and denominator gona vanish

27. ganeshie8

\large{\begin{align} \dfrac{dt}{dx} &= \dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\dfrac{d}{dx}\left(\dfrac{12x}{x^2+108}\right)\\~\\~\\ &=\dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\left(\dfrac{(x^2+108)12-12x(2x)}{(x^2+108)^2}\right)\\~\\~\\ \end{align}} yes ?

28. ganeshie8

setting that equal to 0 gives $\large (x^2+108)12-12x(2x) = 0$ solve $$x$$

29. anonymous

x=10.4

30. anonymous

but sir im a little bit confused by this... how come the remaining equation becomes $(x ^{2}+108)12-12x(2x)$

31. anonymous

how did the others vanish? and this is the equation that still remains?

32. anonymous

where did this $\frac{ 1 }{ 1 +\left( \frac{ 12x }{ x ^{2}+108 } \right)^{2}}$ go?

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